Question

Consider the time series data in columns A and B.

Month (A)	Value (B)
1	6
2	11
3	9
4	14
5	15

a. Construct a time series plot.

b. Use simple linear regression analysis to find the parameters for the line

c. For each new month, how much change is there in the value (on average)?

d. Create a "Forecast" column and use the regression equation to provide a forecast for each month. Include a forecast for month 6.

e. Would you feel comfortable using this equation to provide a forecast for month 18? Explain why or why not.

Answer 1

X	Y	XY	X²	Y²
1	6	6	1	36
2	11	22	4	121
3	9	27	9	81
4	14	56	16	196
5	15	75	25	225

Ʃx =	15
Ʃy =	55
Ʃxy =	186
Ʃx² =	55
Ʃy² =	659
Sample size, n =	5
x̅ = Ʃx/n = 15/5 =	3
y̅ = Ʃy/n = 55/5 =	11
SSxx = Ʃx² - (Ʃx)²/n = 55 - (15)²/5 =	10
SSyy = Ʃy² - (Ʃy)²/n = 659 - (55)²/5 =	54
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 186 - (15)(55)/5 =	21

a) time series plot:

b)

Slope, b = SSxy/SSxx = 21/10 = 2.1

y-intercept, a = y̅ -b* x̅ = 11 - (2.1)*3 = 4.7

Regression equation :

ŷ = 4.7 + (2.1) x

c)

For each new month, the value will increase by 2.1 on average.

d)

Predicted value of y at x = 6

ŷ = 4.7 + (2.1) * 6 = 17.3

e) No, the value is not in the range of values of x. so it would not be reasonable to forecast it.